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Difference between revisions of "2005 AIME II Problems/Problem 10"

(→‎Solution 1)
(→‎Solution 1: The explanation is incorrect; the ratio being asked for is actually the ratio of the volume of the octahedron to the volume of the cube (the larger volume to the smaller volume, thus cannot be less than 1, as 2/9 is).)
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Let the side of the octahedron be of length <math>s</math>.  Let the [[vertex |vertices]] of the octahedron be <math>A, B, C, D, E, F</math> so that <math>A</math> and <math>F</math> are opposite each other and <math>AF = s\sqrt2</math>.  The height of the square pyramid <math>ABCDE</math> is <math>\frac{AF}2 = \frac s{\sqrt2}</math> and so it has volume <math>\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}</math> and the whole octahedron has volume <math>\frac {s^3\sqrt2}3</math>.   
 
Let the side of the octahedron be of length <math>s</math>.  Let the [[vertex |vertices]] of the octahedron be <math>A, B, C, D, E, F</math> so that <math>A</math> and <math>F</math> are opposite each other and <math>AF = s\sqrt2</math>.  The height of the square pyramid <math>ABCDE</math> is <math>\frac{AF}2 = \frac s{\sqrt2}</math> and so it has volume <math>\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}</math> and the whole octahedron has volume <math>\frac {s^3\sqrt2}3</math>.   
  
−
Let <math>M</math> be the midpoint of <math>BC</math>, <math>N</math> be the midpoint of <math>DE</math>, <math>G</math> be the [[centroid]] of <math>\triangle ABC</math> and <math>H</math> be the centroid of <math>\triangle ADE</math>.  Then <math>\triangle AMN \sim \triangle AGH</math> and the symmetry ratio is <math>\frac 23</math> (because the [[triangle median |medians]] of a triangle are trisected by the centroid), so <math>GH = \frac{2}{3}MN = \frac{2s}3</math>.  <math>GH</math> is also a diagonal of the cube, so the cube has side-length <math>\frac{s\sqrt2}3</math> and volume <math>\frac{2s^3\sqrt2}{27}</math>.  The ratio of the volumes is then <math>\frac{\left(\frac{2s^3\sqrt2}{27}\right)}{\left(\frac{s^3\sqrt2}{3}\right)} = \frac29</math> and so the answer is <math>\boxed{011}</math>.
+
Let <math>M</math> be the midpoint of <math>BC</math>, <math>N</math> be the midpoint of <math>DE</math>, <math>G</math> be the [[centroid]] of <math>\triangle ABC</math> and <math>H</math> be the centroid of <math>\triangle ADE</math>.  Then <math>\triangle AMN \sim \triangle AGH</math> and the symmetry ratio is <math>\frac 23</math> (because the [[triangle median |medians]] of a triangle are trisected by the centroid), so <math>GH = \frac{2}{3}MN = \frac{2s}3</math>.  <math>GH</math> is also a diagonal of the cube, so the cube has side-length <math>\frac{s\sqrt2}3</math> and volume <math>\frac{2s^3\sqrt2}{27}</math>.  The ratio of the volumes is then <math>\frac{\left(\frac{s^3\sqrt2}{3}\right)}{\left(\frac{2s^3\sqrt2}{27}\right)} = \frac92</math> and so the answer is <math>\boxed{011}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 19:34, 13 August 2024

Problem

Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$

Solutions

[asy] import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7)); draw(box((-1,-1,-1),(1,1,1))); draw((-3,0,0)--(0,0,3)--(0,-3,0)--(-3,0,0)--(0,0,-3)--(0,-3,0)--(3,0,0)--(0,0,-3)--(0,3,0)--(0,0,3)--(3,0,0)--(0,3,0)--(-3,0,0)); [/asy]

Solution 1

Let the side of the octahedron be of length $s$. Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other and $AF = s\sqrt2$. The height of the square pyramid $ABCDE$ is $\frac{AF}2 = \frac s{\sqrt2}$ and so it has volume $\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}$ and the whole octahedron has volume $\frac {s^3\sqrt2}3$.

Let $M$ be the midpoint of $BC$, $N$ be the midpoint of $DE$, $G$ be the centroid of $\triangle ABC$ and $H$ be the centroid of $\triangle ADE$. Then $\triangle AMN \sim \triangle AGH$ and the symmetry ratio is $\frac 23$ (because the medians of a triangle are trisected by the centroid), so $GH = \frac{2}{3}MN = \frac{2s}3$. $GH$ is also a diagonal of the cube, so the cube has side-length $\frac{s\sqrt2}3$ and volume $\frac{2s^3\sqrt2}{27}$. The ratio of the volumes is then $\frac{\left(\frac{s^3\sqrt2}{3}\right)}{\left(\frac{2s^3\sqrt2}{27}\right)} = \frac92$ and so the answer is $\boxed{011}$.

Solution 2

Let the octahedron have vertices $(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)$. Then the vertices of the cube lie at the centroids of the faces, which have coordinates $(\pm 1, \pm 1, \pm 1)$. The cube has volume 8. The region of the octahedron lying in each octant is a tetrahedron with three edges mutually perpendicular and of length 3. Thus the octahedron has volume $8 \cdot \left(\frac 16 \cdot3^3\right) = 36$, so the ratio is $\frac 8{36} = \frac 29$ and so the answer is $\boxed{011}$.

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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